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Kipish [7]
2 years ago
7

I am standing on the upper deck of the football stadium. I have an egg in my hand. I am going to drop it and you are going to tr

y to catch it. You are standing on the ground. Apparently, you do not want to stand directly under me; in fact, you would like to stand as far to one side as you can so that if I accidentally release it, it won’t hit you on the head. If you can run at 20 feet per second and I am at a height of 100 feet, how far away can you stand and still catch the egg if you start running when I let go?
Engineering
1 answer:
Alina [70]2 years ago
8 0

Answer:

Δx = 25 ft.

Explanation:

Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:

1 ) Egg falling:

If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:

yf-y₀ = 1/2* g* t²

If we take the up direction as positive, we can solve for t as follows:

0-100 ft = 1/2* (-32.15 ft/s²)* t²

⇒ t = \sqrt{(100*2)/32.15} = 2.5 sec.

2) Person on the ground running away:

In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.

We can get the distance at which he can reach, applying the definition of velocity:

v = (xf-x₀) / (tfi-t₀)

If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:

xf = v*t = 20 ft/sec*1.25 sec = 25 ft.

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

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Considering the analogy between electrical circuit and thermal circuit, show your approach to derive an expression for the therm
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Answer:

Thermal resistance for a wall depends on the material, the thickness of the wall and the cross-section area.

Explanation:

Current flow and heat flow are very similar when we are talking about 1-dimensional energy transfer. Attached you can see a picture we can use to describe the heat flow between the ends of the wall. First of all, a temperature difference is required to flow heat from one side to the other, just like voltage is required for current flow.  You can also see that R_{th} represents the thermal resistance. The next image explains more about the parameters which define the value of the thermal resistances which are the following:

  1. Wall Thickness.  More thickness, more thermal resistance.
  2. Material thermal conductivity (unique value for each material). More conductivity, less thermal resistance.
  3. Cross-section Area. More cross-section area, less thermal resistance.

A expression to define  the thermal resistance for the wall is as follows:  R_{th} =\frac{l}{Ak}, where  l is the distance between the tow sides of the wall, that is to say the wall thickness; A is the cross-section area and k is the material conducitivity.

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